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  "body": "\n## Proof\n\n1. Let $\\varphi : G \\to H$ be a group homomorphism with kernel $K$. [[$K$ is a normal subgroup|0C9ZKYSPYR]] of $G$, so the quotient group $G/K$ exists.\n2. Define $\\psi : G/K \\to \\varphi(G)$ by $\\psi(gK) = \\varphi(g)$.\n3. The map $\\psi$ is well defined. If $gK = hK$, then $g^{-1}h \\in K$, so $\\varphi(g)^{-1}\\varphi(h) = e$ and $\\varphi(g) = \\varphi(h)$.\n4. The map $\\psi$ is a homomorphism, because $\\psi(gK\\,hK) = \\psi(ghK) = \\varphi(gh) = \\varphi(g)\\varphi(h)$.\n5. The map $\\psi$ is onto the [[image|0112DTS464]] $\\varphi(G)$, because each $\\varphi(g)$ is $\\psi(gK)$.\n6. If $\\psi(gK) = e$, then $\\varphi(g) = e$, so $g \\in K$ and $gK = K$. Thus the kernel of $\\psi$ is trivial, and [[$\\psi$ is injective|0N2P0XRTDF]].\n7. Thus $\\psi$ is an [[isomorphism|085PND4KNT]], and $G/K \\cong \\varphi(G)$.\n\n## Idea\n\nSend each coset of the kernel to the common image of its elements. This map is well defined, respects products, and has a trivial kernel.\n",
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